Gromov s Proof of Mostow Rigidity

Gromov s Proof of Mostow Rigidity Mostow Rigidity is a theorem about invariants. An invariant assigns a label to whatever mathematical object I am studying that is well-defined up to the appropriate equivalence. For example, the diameter of a metric space is an invariant of metric isometries, since two isometric spaces must have the same diameter. Topological spaces can be assigned homology or homotopy groups. Polyhedra can be assigned their volume. These labels, which may be numbers, groups, or more sophisticated objects, are tremendously useful for distinguishing objects but are rarely exact. Two metric spaces with the same diameter can be very, very far from isometric, for example. A complete invariant is one where the assignment is injective, so that if two objects get assigned the same label they are in fact isomorphic. Of course, complete invariants are very rare. Complicated mathematical structures can almost never be uniquely identified by a simple statistic, even if that statistic is something complex like a group. The problem is, in a sense, that the objects are too flexible to be tied down by any one single restriction. Fix the volume of a polyhedron and you can still wiggle all the angles about, if you do so carefully. What Mostow asserted was that the class of closed, hyperbolic manifolds of dimension n 3 are very rigid, so that there exists a very simple complete invariant. That invariant is, in fact, the fundamental group! Consider how bizarre and incredible this is: if you only tell me what the fundamental group of my manifold looks like I can determine not only its entire topological structure, but even the geometric structure! Underlying every hyperbolic manifold is a simple frame consisting of a set of based loops, and that structure can be completed in a unique way. There are, actually, other excellent invariants for these manifolds: their volume! This has the advantage of being a single number but is not quite complete, although it does determine the hyperbolic manifold up to a finite list. The proof of that is another matter entirely and not the focus of these notes. 1

1 Boundaries at infinity Virtually every proof of Mostow s rigidity goes through the same first step, which is outlined in this section. Take M and N to be closed, hyperbolic manifolds of dimension n 3. Their universal covers are the contractible hyperbolic space H n, so these spaces are K(π, 1) s. What that means is that all their higher homotopy groups vanish. K(π, 1) spaces are uniquely determined up to homotopy equivalence, so we can find a pair of maps f : M N and g : N M that are inverses up to homotopy. If we are careful, f and g can be taken C 1. If we lift these maps to the universal covers equivariantly they remain Lipschitz, and are in fact quasi-isometries. Hyperbolic space H n has a boundary at infinity homemorphic to the sphere S n 1. This boundary at infinity can be thought of the space of geodesics escaping H n up to the fellow traveling equivalence. Since quasiisometries send geodesics to quasi-geodesics that fellow travel a unique geodesic, we can in fact extend our maps to the boundary of hyperbolic space. By abuse of notion I refer to these maps as f and g also. Some tinkering with hyperbolic geometry reveals that these extensions are continuous, and in fact the assignment that takes a map on hyperbolic space to the associated boundary map is injective. What is then the strategy of the proof? We wish to show that the map on the boundary must be conformal. If n 3, the conformal maps on S n 1 all extend isometries on H n, and hence, up to isometry, the map on the boundary is the identity, so the map on the interior is the identity, and thus f and g are isometries. This fails for n = 2 because any diffeomorphism of the circle is conformal, and virtually none of these extend to the interior as isometries. 2

2 Gromov s Proof Gromov s strategy is the following. To show that the boundary map is, up to isometry, the identity, we take a collection of points on the boundary which are the vertex of an ideal simplex of maximal volume. Gromov then claims that these must also be sent to the vertices of an ideal simplex of maximal volume. Up to isometry, we can assume that these points are fixed. Next, reflect one of these vertices through the face determined by the others and consider where this new vertex is sent. This new vertex, together with the vertices of that face, give a new simplex which also has maximal volume (because it is regular, a separate result). This must again be sent to the vertices of an ideal simplex of maximal volume, that is the vertices of an ideal regular simplex, but because we have fixed all but one of the vertices and because the boundary map is injective the newest vertex must also be fixed. Repeating this we see that a whole mesh of the boundary is fixed, and if we take reflections carefully we can show that a smaller mesh is fixed, and so on, so that a dense mesh if fixed. By continuity, the boundary map is the identity. So how do we justify this claim that the vertices of an ideal simplex of maximal volume must be sent to the vertices of an ideal simplex of maximal volume? Put another way, how does Gromov make arguments about volume simply from the topological information that the maps f and g are homotopy equivalences? He uses a norm on homology that encodes volumetric information. 2.1 The Gromov Norm For any topological space X let C (X) be the real singular chain complex. Any k-chain can be written uniquely as c = i a i σ i We can endow our chain complex with the l 1 norm by setting c = i a i. This descends to homology as a seminorm by taking the infimum of the norms in the homology class. Note that if f : X Y is a continuous map and z is a homology class that 3

f (z) z Because the induced map on homology can collapse certain simplices on to each other. Now, as M and N are orientable they have fundamental classes [M] and [N]. The quantity [M] is called the Gromov norm of the manifold, or the simplical volume. In general, the Gromov norm of a manifold may be zero, but the Gromov norm of hyperbolic manifolds will be positive, and in fact related to their volume. Before we go foreward we will define an operation on simplices that normalizes them. Given a simplex σ in M we lift it to the universal cover (thought of as the hyperboloid model in R n+1 ), build the affine simplex in the ambient Euclidean space, project it back down to the hyperboloid, and then let that descend to M. This is called the straightening of σ, written str(σ) and it induces a well-defined map on homology. Let v n be the supremum of the volumes of straight n-simplices in H n. We claim that this is in fact bounded. To be precise, for n 2, v n π (n 1)! For n = 2 this is Gauss-Bonnet. In higher dimensions one proves that v n v n 1. This is derived via an explicit calculation and some convenient n 1 normalizations. The next stop in the proof is to justify the claim that the Gromov norm is related to the volume. The exact relation is M = Vol(M) v n That is, if we scale the simplical norm by the maximum possible norm for straight simplices we get exactly the volume of M. This is quite the incredible claim, and the proof is a little involved. It turns out that it is convenient to replace simplical homology with an isomorphic, measure-theoretic homology theory. In this new theory, the k-chains are signed Borel measures on C 1 ( k, M) with bounded total variation and compact support. We denote these chains by C k (M). 4

What is the boundary map on these chains? Well, on ordinary chains we can take projection onto the ith face. This induces a pushforward on the space of measures, where the ith projection of a measure on k-chain takes a collection of (k 1)-chains, finds the collection of k which i-project to it, and measures those. The alternating sum of these projections gives the differential, and it is not hard to see that we get a chain complex, i.e. the composition of differentials is zero. There is a natural chain map from our original chain complex to this new one. It sends a chain σ to the Dirac measure on σ. This descends to homology and one can check that this gives an isomorphism. The usual DeRham pairing of k-chains and k-forms extends to a pairing with the k- chains in C k (M), via ( ) (µ, α) τdα dµ(τ) τ C 1 ( k,m) k Put in works, integrate the form over all possible k-chains, summing the contributions via the weighting of the measure. We are now ready to prove that vol(m) v n M. The volume of M can be obtained by pairing the fundamental class [M] with the volume form Ω M. Take a generator for the fundamental class composed only of small, straight simplices. Via the isomorphism between our homology theories we can replace the fundamental class with a fundamental measure µ. Now, the straightening operation is also defined on measures, simply by stating that the straightened measure of a simplex is the measure of the straightened simplex. Since we took the fundamental class to be straight we also have that µ is straight, so it only puts mass on straight simplices. Thus we just end up integrating the form against straight simplices. Let τ be such a simplex. The pullback τ Ω M is equal to τ Ω H n for τ a lift of τ. Each of those integrals can have measure at most v n by definition, so a simple ML-estimate tells us that the whole integral is bounded by v n times the mass of µ. Taking the infimum over representaties of [M], one obtains vol(m) v n M. This works because every representative for [M] can be straightened (and this just shrinks the norm). This proves that the norm of M, which can be obtained by integrating a volume form over a fundamental class (taken here to be straight), is bounded by the size of that fundamental 5

class times the largest possible volume of each of its pieces. Now to show that the vol(m) v n M, we will explicitly construct a cycle whose norm is Vol(M)/v n. First, recall that Isom + (H n ) is a unimodular Lie group. If we let Γ = π 1 (M) and denote the quotient Isom + (H n )/Γ by G, then since Γ was discrete G is again unimodular, with a Haar measure h G. We can normalize this measure to have norm Vol(M). Fix σ a simplex in H n. We get a map ψ σ from G into C 1 ( n, M) given by hitting σ with an element of G and projecting it down to M. This defines a map smear(σ) from C 1 ( n, H n ) into C n (M) which sends σ to the pushforward measure ψ σ (h G ). This measure gives all of its support to projected G-shifts of σ, and given a collection of such simplices assigns them value that h M would assign the associated shifts in G. The smear map has a number of nice properties. For one thing, σ and gσ have the same smear. If we take the ith face of σ and smear that it is equivalent to smearing σ and taking the ith face of the resulting measure. The norm of smear(σ) is just the volume of M because of our normalization. Lastly, pairing smear(σ) with the volume form Ω M gives Vol(M) Vol(σ). To see this last fact, observe that this pairing only weights the integrals of the volume form over projected G-translated of σ, all of which are equal to n σ Ω H n = Vol(σ) We d like to take this smeared measure as our witness for a cycle that hits the floor of Vol(M)/v n provided by our earlier bound. The only problem is that it is not a cycle. However, if we take σ and reflect it through one of its faces to get σ, and then write ζ(σ) = 1 2 (smear(σ) smear(σ )) then this is indeed a cycle, but it is non trivial. Moreover, this measure has norm equal to Vol(M). Since ζ(σ) is a top-dimensional measure it must be equal to [M] for some λ. We already know that Vol(σ) Vol(M) = ζ(σ), Ω M = λ [M], Ω M = λ Vol(M) Thus λ = Vol(σ). We also know that 6

Vol(M) = ζ(σ) Vol(σ) M Taking the supremum over all straight simplices gives Vol(M) v n M. Put into words, we built a measure whose norm is exactly the volume of M and which represents the class of Vol(σ)[M]. 3 The Proof So far we ve seen that the Gromov norm of the fundamental class is exactly proportional to the volume. What remains to be shown is our claim about sending vertices of ideal maximal volume simplices to vertices of ideal maximal volume simplices. Take v 0,, v n to be the vertices of an ideal simplex of maximal volume σ. Suppose that the volume of the straightening of h(σ) is not maximal. Then I could wiggle around the v i within open sets U i H n and keep the volume of the straightened image bounded away from v n by 2ɛ, for some ɛ > 0. Within each U i one can take V i, open subsets, such that the set A(G) = g Isom + (H n ) : v i V i gv i U i } has positive measure m A > 0. However, for any δ > 0 we can find some wiggled simplex σ 0. with volume at least v n δ. Let s smear this to obtain a measure on the n-chains of M, push it forward to N and straighten it there. What do we get if we pair this with Ω N? By definition we integrate over all n-chains in N, pairing each chain with the volume form. By definition of the pushforward measure this is just the integral over G with its Haar measure of the volume of the pullback of Ω N through the f-pushforward of gσ 0. Now this integral splits into two pieces. On A(G), which has measure m A, this is the integral of the volume form of H n over the straightened fgσ 0, which by definition is less than v n 2ɛ, and certainly less than v n 2ɛ + δ. When integrating over the rest of G the integral of the volume form over the pushforward is no bigger than v n which is less than Vol(σ 0 ) + δ. This is, all in all m A (Vol(σ 0 ) 2ɛ + δ) + (Vol(M) m A )(Vol(σ 0 ) + δ) 7

which simplifies to Vol(M)(Vol(σ 0 ) + δ) 2m A ɛ If we take δ < (ɛm A )/ Vol(M) then this is < Vol(M) Vol(σ 0 ) ɛm A However, we know f : M N is a homotopy equivalence, sending [M] to [N], so M and N must have the same volume. Since ζ(σ 0 ) represents Vol(σ 0 )[M] its straightened pushforward must represent Vol(σ 0 )[N]. However, the above calculation suggests that the pushforward must represent λ[n] with λ < Vol(σ 0 ) ɛm A Vol(M) This gives a contradiction. To summarize this argument, since f is a homotopy equivalence it must send fundamental classes to fundamental classes, preserving volume, sending representatives of λ[m] to representatives of λ[n]. But if h does not preserve collections of maximal volume vertices on the boundary one can find a simplex whose smear represents λ[m] and whose image represents ρ[n] for some ρ < λ. 8